Bending Deflection –Statically Indeterminate BeamsAE1108-II: Aerospace Mechanics of Materials

Aerospace Structures& Materials

Faculty of Aerospace Engineering

Dr. Calvin RansDr. Sofia Teixeira De Freitas

Recap

I. Free Body Diagram II. Equilibrium of Forces (and Moments)III. Displacement CompatibilityIV. Force-Displacement (Stress-Strain) Relations

V. Answer the Question! – Typically calculate desired internal stresses, relevant displacements, or failure criteria

Procedure for Statically Indeterminate Problems

Solve when number of equations = number of unknowns

For bending, Force-Displacement relationships come from Moment-Curvature relationship

(ie: use Method of Integration or Method of Superposition)

Statically Indeterminate BeamsMany more redundancies are possible for beams:

- Draw FBD and count number of redundancies- Each redundancy gives rise to the need for acompatibility equation

P

A B

P

VA VB

HAMA

- 4 reactions- 3 equilibrium equations 4 – 3 = 1 1st degree statically indeterminate

Statically Indeterminate BeamsMany more redundancies are possible for beams:

- Draw FBD and count number of redundancies- Each redundancy gives rise to the need for acompatibility equation

- 6 reactions- 3 equilibrium equations 6 – 3 = 3 3rd degree statically indeterminate

P

A B

P

VA VB

HAMA HB

MB

Solving statically indeterminate beams using method of integration

What is the difference between a support and a force?

F

Displacement Compatibility (support places constraint on deformation)

Method of IntegrationP

A B

P

A B

VB

What if we remove all redundancies and replace with reaction forces?

If we treat VBas known, we

can solve!

Can integrate to find v

2

2

d vM EIdz

Formulate expression for M(z)

Method of Integration (cont)P

A B

VB

2

2 , Bd vEI M z Vdz

, BEIv M z V dz

( , )BEIv M z V dz

C1

C2

Determine constants of integration from Boundary Conditions

A

At A, v = 0, θ = 0

( , )Bv z V How to determine VB?

Method of Integration (cont)P

A B

VB

( , )Bv z V How to determine VB?

P

A B=

Compatibility BC: At B, v = 0

VB2PVB1

VB changes displacement!

Must be compatible!

Solve for VB

Compatibility equations for beams are simply the boundary conditions at redundant supports

11Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 1Problem Statement

q

A BDetermine deflection equation for the beam using method of integration:

Treat reaction forces as knowns!

0 AF H

2) Equilibrium: 1) FBD:

A BVA VB

HAMA

q

2

2 AqL LV

Solution

A BF V V qL

2

2A A BqLM M LV

12Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 1

4) Determine moment equation:

AVA

V

MA

q

M

z

ccwzM M

2

2A AqM V z z

Can also use step function approach

2zqzAM AV z

M 00AM z 0AV z 202q z BV z L

always off

always on

13Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 1

2 312 6

AA

V qEIv M z z z C

5) Integrate Moment equation to get v’ and v

2

2A AqEIv M V z z = M(z)

= -EIθ(z)

2 3 4

1 22 6 24A AM V qEIv z z z C z C = -EIv(z)

q

A B

We now have expressions for v and v’, but need to determine constants of integration and unknown reactions

z

14Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 1

5a) Solve for Constants of Integration using BC’s:

= -EIθ(z)

= -EIv(z)

q

A B

z

Boundary Conditions:At z = 0, θ = 0 3 2

10 0 0 06 2

AA

VqEI M C

1 0C

0

0Fixed supportθ = 0, v = 0

2 312 6

AA

V qEIv M z z z C

2 3 41 22 6 24

A AM V qEIv z z z C z C

15Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 1

5a) Solve for Constants of Integration using BC’s:

= -EIθ(z)

= -EIv(z)

q

A B

z

Boundary Conditions:

0

0Fixed supportθ = 0, v = 0

2 312 6

AA

V qEIv M z z z C

2 3 41 22 6 24

A AM V qEIv z z z C z C

At z = 0, v = 0 4 3 220 0 0 0

24 6 2A AV MqEI C

2 0C

0

16Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 1

5b) Solve for Reaction Forces using BC’s (imposed by redundant support):

= -EIθ(z)

= -EIv(z)

q

A B

z

Boundary Conditions:

roller supportv = 0

2 3

2 6A

AV qEIv M z z z

2 3 4

2 6 24A AM V qEIv z z z

At z = L, v = 0 4 3 2024 6 2

A AV MqEI L L L

34

AA

MqLVL

Recall from equilibrium:

2

2A AqLM LV B AV qL V

17Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 1

5b) Solve for Reaction Forces using BC’s (imposed by redundant support):

q

A B

z

23 , , 4 2

AA A A B A

MqL qLV M LV V qL VL

38BV qL

58AV qL

2

8AqLM A B

VA VB

MA

18Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 1

We were asked to determine deflection equation:

q

A B

z

2

2 23 5 248qzv L Lz z

EI

0 0.5 10.4

0.2

0

LMax Displacement:

2 2 36 15 8 048

qv L z Lz zEI

0.5785z L

=0

2

(0.5785 ) 0.005416 qLv LEI

2 3 4

2 6 24A AM V qEIv z z z

19Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 1

Now that the reactions are known: q

A B

z

5( )8qLV z EIv qz

2( )2A AqM z EIv M V z z

2 258 8 2

qL qLz qz

ddz

M4L

2

8qL

V58

qL

38

qL58

L

Solving statically indeterminate beams using superposition

Method of Superposition

P

A B

P

VA VB

HAMA

- 4 reactions- 3 equilibrium equations

1st degree statically indeterminate

0

3 2

A

A B

A B A

F H

F V V P

M LV LP M

2) Equilibrium:

1) FBD:2L L

Determine reaction forces:

Method of Superposition (cont)How do we get compatibility equation?

P

A B

P

A B

VB

A BδB1

δB2

Split into two statically determinate problems

3) Compatibility: δB1 = δB2 1 2 0B Bv v Be careful!+ve v

Method of Superposition (cont)How do we get Force-Displacement relations?

Can integrate to find v

2

2

d vM EIdz

We have been doing this in the previous lectures

Integrate Moment Curvature Relation Standard Case Solutions

P

A BL

2

2BPLEI

3

3BPLvEI

+ +

Method of Superposition (cont)From the standard case:

P

A B δB1

P

A BL

2

2BPLEI

3

3BPLvEI

2L L 2 22 22P

P L PLEI EI

1B P Pv v L

3143PLEI

straight

4) Force-Displacement:

3 22 23

P L PL LEI EI

Method of Superposition (cont)From the standard case:

VB

A B

δB2 3 3

2

3 93B B

B

V L V LvEI EI

Compatibility:

1 2 0B Bv v 33 914

3BV LPL

EI EI

1427BV P

4) Force-Displacement:

P

A BL

2

2BPLEI

3

3BPLvEI

Additional remarks about bending deflections

Remarks about Beam DeflectionsRecall Shear deformation

Moment deformation+

Negligible (for long beams)

Bending Deformation = Shear Deformation + Moment Deformation

+

M M

+ V

V

V(x)

M(x)

Remarks about Beam Deflections

A B

C

P

P

A

P

BVB

HB

MB

B

C

P

VB HB

MB

AB

C

P

PAxial deformation of AB due to HB

Axial deformation of BC due to VB

Axial deformation << bending deformation!

Remark about Beam Deflections

negligibleDeformation = Axial Deformation + Shear Deformation + Moment Deformation

For bending deformation problems

A

P

BVB

HB

MBBUT!If moment deformation is not

present, deformation is not negligible

30Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 2

A

q

B

D

CCalculate reaction forces at A and D:

First, can we see any simplifications?

Symmetry!

L, EI

Symmetry implies: - Reactions at A = reactions at D- Slope at symmetry plane = 0- Shear force at symmetry plane = 0

We will solve using superposition and standard cases A

BHF MF

q

31Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 2Calculate reaction forces:

A

BHF MF

q

VAHA

MA

1) FBD

2) Equilibrium

A FF H H

2AqLF V

2

8ccwA A F F

qLM M H L M

F

Split problem into two: beam AB and beam BF

32Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 2Calculate reaction forces:

HF

F

A

BHF MF

q

=

A

B

MF - qL2/8

qL/2

+

q

B

HF MF

F

MF - qL2/8

HF

qL/2

3) Compatibility

F 0, vB 0(axial deformation negligible)

33Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 24a) Force-Displacement for Beam AB

HF

A

B

MF - qL2/8

qL/2

HF

A

B

A

BqL/2

A

B

MF - qL2/8

3

3F

BH Lv

EI

2

2cw FB

H LEI

0Bv

0B

2 4

2 16F

BM L qLv

EI EI

3

8cw FB

M L qLEI EI

+ +

34Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 24a) Force-Displacement for Beam AB

HF

A

B

MF - qL2/8

qL/2 3 2 4

3 2 16F F

BH L M L qLv

EI EI EI

2 3

2 8cw F FB

H L M L qLEI EI EI

0, v 0F B Recall compatibility:

0

Still need θF

35Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 24b) Force-Displacement for Beam BF

326

cwF

q LEI

0F cw FF

M LEI

q

B

HF MF

F

MF + qL2/8

HF

qL/2 L/2

B

HF

F B

MF

FB F

q+ +

3

48cw FF

M LqLEI EI

Wait! Not entirely correct!

36Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 2

3

48cw FF B

M LqLEI EI

Previous angle relative to fixed support B

q

B

HF MF

F

θB

23 2748 2

cw F FF

M L H LqLEI EI EI

0

37Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 2

0, v 0F B

23 27 048 2

cw F FF

M L H LqLEI EI EI

Solve:

Recall compatibility:

3 2 4

03 2 16

F FB

H L M L qLvEI EI EI

2

8

24

F

F

qLH

qLM

A FF H H

2AqLF V

2

8ccwA A F F

qLM M H L M

Recall equilibrium:

2

82

3

A

A

qLH

qLM

38Aerospace Mechanics of Materials (AE1108-II) – Example Problem

Example 2

A

BHF

MF

VAHA

MA

F

A

q

B

D

C

L, EI

2

8

24

F

F

qLH

qLM

2

8

22

3

A

A

A

qLH

qLV

qLM

For next time